calculate the conditional distribution of brownian motion

Suppose $W=(W_t)$ is a Brownian Motion with respect to a filtration $(\mathcal{F}_t)$. How can I compute the conditional distribution of $W_{t+h}$ given $\mathcal{F}_t$.

I started like this: $W_{t+h}-W_t$ is idependent of $\mathcal{F}_t$ and normald distributed with mean $0$ and variance $h$. Then I wrote $W_{t+h}=W_t+ (W_{t+h}-W_t)$, hence I have to compute:

$$P[W_{t+h}=W_t+ (W_{t+h}-W_t)\in A|\mathcal{F}_t]$$

For $A\in \mathcal{B}(\mathbb{R})$. I wrote the conditional probability as a expectation of an indicator function. The result should be a normal distribution with mean $W_t$ and variance $h$. Thanks for your help

math

• I'm confused. Doesn't the martingale property give $\mathbb E[W_{t+h}|\mathcal F_t]=W_t$? – user31373 Jun 20 '12 at 23:44
• @ Leonid Kovalev: Yes, of course, but writing the above conditional probability as an expectation, leads to: $P[W_{t+h}=W_t+(W_{t+h}-W_t)\in A|\mathcal{F}_t]=E[\mathbf1\{W_{t+h}=W_t+(W_{t+h}-W_t)\in A\}|\mathcal{F}_t]$. I do not see how to proceed from here – math Jun 21 '12 at 6:50

The conditional distribution of $W_{t+h}$ conditionally on $\mathcal F_t$ is a random distribution, that is, a map $M:\Omega\to\mathcal M_1^+(\mathbb R,\mathcal B(\mathbb R))$, measurable with respect to $\mathcal F_t$, and such that, for every bounded measurable function $u$, $$\mathrm E(u(W_{t+h})\mid\mathcal F_t)=\int_{\mathbb R} u\mathrm dM\quad\text{almost surely}.$$ The OP explains why $$\mathrm E(u(W_{t+h})\mid\mathcal F_t)=\mathrm E(u(W_{t}+Z_h)\mid W_t)\quad\text{almost surely},$$ where $Z_h$ is centered normal with variance $h$ and independent of $W_t$. Thus, for every bounded measurable function $u$, $$\int_{\mathbb R} u\mathrm dM=\int_{\mathbb R} u(W_t+z)\mathrm d\gamma_h(z)\quad\text{almost surely},$$ where $\gamma_h$ is the centered normal distribution with variance $h$. This proves that, for $\mathrm P$-almost every $\omega$, the distribution $M(\omega)$ is normal with mean $W_t(\omega)$ and variance $h$.
• @Did could you give more details about the last step? Why can can you conclude it carring only about the distribution of $Z$? – sky90 Jun 22 '15 at 13:06
• @sky90 Because $W_{t+h}$ is distributed as $W_t+Z_h$ where $W_t$ is $\mathcal F_t$-measurable and $Z_h$ is independent of $\mathcal F_t$, and because, for every $w$, $w+Z_h$ is normal with mean $w$ and variance $h$. – Did Jun 22 '15 at 15:31